The sum of three consecutive integers is multiplied by 33. That result is equivalent to the product of the integers. What are the integers, their sum, and their product?
The integers are 9, 10, and 11. The sum of these integers is 30. Their product is 990.
(9x10x11=990, 9+10+11=30 30x33=990)
Let the consecutive numbers = n, n+1, n+2,
33[n+ n+1+ n+2] = n* (n+1)* (n+2), solve for n
Solve for n:
33 (3 n+3) = n (n+1) (n+2)
Expand out terms of the left hand side:
99 n+99 = n (n+1) (n+2)
Expand out terms of the right hand side:
99 n+99 = n^3+3 n^2+2 n
Subtract n^3+3 n^2+2 n from both sides:
-n^3-3 n^2+97 n+99 = 0
The left hand side factors into a product with four terms:
-((n-9) (n+1) (n+11)) = 0
Multiply both sides by -1:
(n-9) (n+1) (n+11) = 0
Split into three equations:
n-9 = 0 or n+1 = 0 or n+11 = 0
Add 9 to both sides:
n = 9 or n+1 = 0 or n+11 = 0
Subtract 1 from both sides:
n = 9 or n = -1 or n+11 = 0
Subtract 11 from both sides:
Answer: |n = 9 and 10 and 11
